do we have y = √(x+8) ?
I will assume that, if (8,4) lies on it
slope
= lim ( √(16+h) - 4)/h , as h --->0
multiply top and bottom by (√(16+h) + 4)
= lim ( √(16+h) - 4)/h * (√(16+h) + 4)/(√(16+h) + 4)
= lim (16 + h - 16)/(h(√(16+h) + 4))
= lim h/(h(√(16+h) + 4))
= lim 1/(√(16+h) + 4) , as h --->0
= 1/(4+4)
= 1/8
or in general:
slope =
Lim ( √(x+h + 8) - √(x+8) )/h , as h ---> 0
multiply top and bottom by ( √(x+h + 8) + √(x+8) )
= Lim ( √(x+h + 8) - √(x+8) )/h * ( √(x+h + 8) + √(x+8) ) / ( √(x+h + 8) + √(x+8) )
= lim (x+h + 8 - (x+8) / (h( √(x+h + 8) + √(x+8) ))
= lim h/(h( √(x+h + 8) + √(x+8) ))
= lim 1/( √(x+h + 8) + √(x+8) ) , as h --->0
= 1/(√(x+8) + √(x+8) )
= 1/ (2√(x+8))
at (8,4) we get
1/(2√16) = 1/8
use the limit process to find the slope of the graph of 'square root' x+8 at (8,4)
a. 4
b. 1/4
c. the slope is undefined at this point
d. 1/8
e. infinity
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